Question

lead has specific heat of .028kcal/kgC, melting point of 328 degrees C and a heat fusion of 5.5 kcal/kg. how much heat must be provided to melt a 250kg sample of lead with a temperature of 20 degrees Celsius?

Answer 1

Q = M*[C*(delta T) + H]

detlta T = 308 C
H = 5.5 kcal/kg
C = 0.028 kcal.kg*C
M = 250 kg

Do the calculation for the heat (Q) in kcal

Answer 2

to get the heat released or absorbed,
Q = mc(T2-T1)
where
m = mass of substance
c = specific heat capacity
T2 = final temperature
T1 = initial temperature
**note: if Q is (-), heat is released and if (+), heat is absorbed

now we can only apply this to substances that did not change its phase, but in the problem, we see that phase change occurs. from solid lead->liquid lead.
thus we need another data called Latent Heat of Fusion to calculate for the heat required to change its phase:
H = m(Lf)

where
m = mass
Lf = Latent Heat of Fusion (fusion means melting)

substituting:
Q = mc(T2-T1)
Q = 250*(0.028)*(328-20)
Q = 2156 cal
*note that is is only for the temp change,, it's phase is still solid.
now to change its phase,
H = m*Lf
H = 250*5.5
H = 1375 cal

thus the total heat (Q,total) needed is
Q,total = Q + H = 2156 + 1375
Q,total = 3531 calories

hope this helps~ :)